$\mathcal{L}(\cos bt}) = \dfrac{s}{s^2 + b^2}$
$\mathcal{L}(\cos t}) = \dfrac{s}{s^2 + 1^2}$
$\mathcal{L}(\cos t}) = \dfrac{s}{s^2 + 1}$
Thus,
$\mathcal{L}(e^{-3t} \cos t}) = \dfrac{s + 3}{(s + 3)^2 + 1}$
$\mathcal{L}(e^{-3t} \cos t}) = \dfrac{s + 3}{(s^2 + 6s + 9) + 1}$
$\mathcal{L}(e^{-3t} \cos t}) = \dfrac{s + 3}{s^2 + 6s + 10}$ answer